Keywords

MSC

39A1405A10

1 Introduction

Let \(\mathbb{N}\) denote the set of natural numbers and \(\mathbb{N} _{0}=\mathbb{N}\cup \{0\}\). Let \(k,l\in \mathbb{N}_{0}\) be such that \(k< l\), then the notation \(j=\overline{k,l}\) means \(k\le j\le l\). In the rest of this section we give some motivation for the study, as well as notions that will be used in the rest of the paper.

while for \(k=0\) a direct calculation shows that \(\Delta x^{(0)}=0\).

The set of polynomials \(\{x^{(k)}: k\in \mathbb{N}_{0}\}\), is linearly independent. Indeed, assume

$$ \sum_{j=0}^{n}a_{j}x^{(j)}\equiv 0, $$

(2)

for some real numbers \(a_{j}\), \(j=\overline{0,n}\), and for a fixed but arbitrary \(n\in \mathbb{N}_{0}\). Then by acting with operator Δ on both sides of equality (2) n times, and using (1), we get \(n!a_{n}=\sum_{j=0}^{n}a_{j}\Delta^{(n)} x^{(j)} \equiv 0\), from which it follows that \(a_{n}=0\). Hence \(\sum_{j=0} ^{n-1}a_{j}x^{(j)}\equiv \sum_{j=0}^{n}a_{j}x^{(j)}\equiv 0\). By repeated use of the argument we get \(a_{n-1}=\cdots =a_{1}=a_{0}=0\), from which the claim follows. From this it easily follows that the set \(\{x^{(k)}: k\in \mathbb{N}_{0}\}\) is a basis in the linear space of all polynomials, so that

The relations \(x=x^{(1)}\), \(x^{2}=x^{(2)}+x^{(1)}\) and \(x^{3}=x^{(3)}+3x ^{(2)}+x^{(1)}\), suggest that \(s^{n}_{0}=0\), \(n\in \mathbb{N}\) (note that \(s^{0}_{0}=s^{0}_{0}x^{(0)}=x^{0}=1\)). Indeed, assume that we have proved \(s^{j}_{0}=0\), \(j=\overline{1,n}\) for some \(n\in \mathbb{N}\). Then, since \(x^{n+1}=x^{n} x\) and \(xx^{(k)} = x^{(k+1)} + kx^{(k)}\), \(k\in \mathbb{N}_{0}\), by using the hypothesis and (3), we get

Comparing the coefficients in (4) we get \(s^{n+1}_{0}=0\), from which the statement follows. Beside this, the following recurrent relation holds:

$$\begin{aligned} s^{n+1}_{k}=s^{n}_{k-1}+ks^{n}_{k}, \end{aligned}$$

(5)

when \(2\le k\le n\), and that \(s^{n+1}_{n+1}=s^{n}_{n}\). From this equality and since \(s^{0}_{0}=s_{1}^{1}=1\) we get \(s^{n}_{n}=1\) for \(n\in \mathbb{N}_{0}\). On the other hand, from \(s^{j}_{0}=0\) for \(j=\overline{1,n}\) and (5) we get \(s^{n+1}_{1}=s^{n}_{1}\), which along with \(s^{1}_{1}=1\) implies that \(s^{n}_{1}=1\) for \(n\in \mathbb{N}\). The numbers \(s^{n}_{k}\), \(n,k\in \mathbb{N}\) are called Stirling’s numbers of the second kind, and the above described procedure is one of the ways how these numbers can be obtained (see, for example, [1, 2]). Another classical approach in getting these numbers is combinatorial. Namely, \(s^{n}_{k}\) represents the number of ways to partition a set of n elements into k nonempty subsets (for details see, for example, [3]). Stirling numbers can be calculated explicitly. In [3] are given several explicit formulas and probably the nicest one is the following:

For more on the Stirling numbers and related topics, see, for example, [1, 3, 4].

Old topic of solving difference equations and systems of difference equations with their applications (see, for example, [1, 2, 4–8]), has re-attracted some recent attention. The publication of our note [9], in which we explained how a nonlinear difference equation can be solved in an elegant way, by transforming it to a linear, has attracted some attention. The results and methods in [9] were later extended for the case of higher-order difference equations in [10], and for some related systems in [11] and [12]. The main idea from [9] have been used and developed a lot in many other papers (see, for example, [13–19] and numerous references therein). One of the joint features in these papers is that the equations/systems therein are somehow transformed to (solvable) linear ones. A frequent situation is that the following difference equation:

where \((b_{n})_{n\in \mathbb{N}}\), \((c_{n})_{n\in \mathbb{N}}\), and \(a_{0}\) are real or complex numbers, decides the solvability. Some methods for solving equation (8) can be found in [2, 6] (see, also the introduction in [17], the references therein, as well as numerous special cases of the equation appearing therein). For some results on the solvability of systems of difference equations see, for example, [12, 14, 16, 18–20]. Some related, but considerably different methods (see, for example, [21, 22] and the related references therein) are used for solving product-type equations/systems.

This means that the Stirling numbers are a solution to a two-dimensional recurrent relation with given boundary conditions \(s^{n}_{n}\) and \(s^{n}_{0}\), \(n\in \mathbb{N}_{0}\). The fact that equation (9) has a closed form formula for the solution with the conditions \(d_{n,0}=0\), \(d_{n,n}=1\), \(n\in \mathbb{N}\), suggests that (5) can be solved on the following domain \(A=\{0\le k\le n; n,k\in \mathbb{N}_{0}\}\), which we call the combinatorial domain. Some classical methods for solving partial difference equations can be found, for example, in [1] and [8], whereas some recent ones can be found, for example, in [23–28], but an analysis shows that they are not suitable for getting a formula for solutions to (9) for a special shape of domain A.

Motivated by [29] and numerous recent applications of equation (8) (for example, those in [9, 10, 13–17, 19, 20]), here we show that there is a closed form formula for solutions to an extension of equation (9) on domain \(A\setminus \{(0,0)\}\) in terms of given boundary values \(d_{n,0}\) and \(d_{n,n}\), \(n\in \mathbb{N}\).

2 Main results

We show how general solution to (9) on set \(A\setminus \{(0,0) \}\) can be found by using a method in [29], which we call the method of half-lines. Namely, the domain is divided into some half-lines, and (9) is regarded on each line as an equation of type (8). It is solved, and based on the obtained formulas one gets the general solution. However, the method cannot be applied directly, so it needs some modifications.

Using the change of variables \(x_{k}=d_{k+1,k}\) equation (10) can be seen as a special case of equation (8), since \(d_{k,k}\) can be regarded as an ‘independent’ variable due to the fact that the multi-index \((k,k)\) belongs to the boundary of \(A\setminus \{(0,0)\}\). In fact, since in (10) the corresponding coefficients \(b_{k}\) are equal to one, here we use the telescoping method of summation (this is a specificity for the equation but the method can be applied for other values of the coefficients too) and get

Theorem 2 shows that the calculation of iterated sums plays an important role in solving equation (23). For the equation treated in [29] we have \(f(k)=1\) for every \(k\in \mathbb{N}\), so that the iterated sums in (25) becomes

This value of the sum is known and is equal to \(C^{k+l-s}_{l-s+1}\) (see, for example, [2]), which is one of the main reasons why in this case we have a closed form formula for its general solution. The process of calculating the sum was essentially explained in [29].

Now we will apply the method described above to the following partial difference equation:

$$\begin{aligned} d_{n,k}=d_{n-1,k-1}+z^{k}d_{n-1,k}, \end{aligned}$$

(26)

where z is a complex number different from 1.

By doing this we see that the following relation, corresponding to equation (12), holds:

Proof

By choosing \(l=n-k\) in (33), using the conditions in (34), and some calculations, equation (35) is easily obtained. □

Declarations

Acknowledgements

The author would like to express his sincere thanks to the referee of the paper for her/his comments and for suggesting some references on the solvability of partial difference equations.

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Competing interests

The author declares that he has no competing interests.

Author’s contributions

The author contributed alone to the writing of this paper. He read and approved the manuscript.

Authors’ Affiliations

(1)

Mathematical Institute of the Serbian Academy of Sciences

(2)

Operator Theory and Applications Research Group, Department of Mathematics, King Abdulaziz University